Решите систему неравенств

(4^x + 5) \ (2^x - 11) >= -1
2^(2x) + 5 >= - 2^x + 11
(2^x)^2 + 2^x - 6 >= 0 -------> 2^x = t
t^2 + t - 6 >= 0 ----------------> (t + 3)(t - 2) >= 0 -----> t1 =< -3 или t2 >= 2 =>
2^x = t1 -----> 2^x =< -3 - быть не может (2 в любой степени > 0)
2^x = t2 -----> 2^x >= 2 -----> 2^x >= 2^1 -----------------------------------> x >= 1

6*log(2x) x + 2*log(4Vx) 2x >= 1 -------> в дальнейшем замена ---> log(2) x = t
___ log(2x) x =
= 1 \ log(x) 2x = 1 \ [log(x) 2 + log(x) x] = 1 \ [log(x) 2 + 1] =
= 1 \ [1 \ log(2) x + 1] = 1 \ [(1 + log(2) x) \ log(2) x] =
= log(2) x \ (1 + log2 x) = t \ (1 + t)

___ log(4Vx) 2x =
= 1 \ log(2x) 4Vx = 1 \ [log(2x) 4 + log(2x) Vx] = 1 \ [log(2x) 2^2 + log(2x) x^(1\2)] =
= 1 \ [2 * log(2x) 2 + 1\2 * log(2x) x] ------------------------> (1)
Преобразования получившихся логарифмов:
_______ log(2x) 2 = 1\ log(2) 2x = 1\ [log(2) 2 + log(2) x] = 1\ [1 + log(2) x] = 1\ (1 + t)
____________________ log(2x) x = 1 \ [log(x) 2x] = 1 \ [log(x) 2 + log(x) x] =
= 1 \ [log(x) 2 +1] = 1 \ [1\log(2) x) + 1] = 1 \ [1\t + 1] = 1 \ [(1 + t) \ t] = t \ (1 + t)
Подставить в (1) результаты:
= 1 \ [2 * 1\(1 + t) + 1\2 * t\(1 + t)] =
= 1 \ [2\(1 + t) + t\2*(1 + t)] =
= 1 \ [(4 + t) \ 2*(1 + t)] = 2*(1 + t) \ (4 + t)
в условие:
6* t\(1+t) + 2*2(1+t)\(4t) >= 1
Дальше легко